# Focal length question

1. Aug 31, 2010

### Sarial

1. The problem statement, all variables and given/known data
An object is placed in front of a concave mirror, 15.0 cm from the mirror's focal point. The image formed by the mirror is five times farther away from the focal point. Calculate the focal length of the mirror.

There are actually two possible image distances di that satisfy the statement of the problem. Enter the larger of the two image distances.

Enter the smaller of the two image distances.

2. Relevant equations
di/do = magnification
1/di+1/do=1/f

3. The attempt at a solution

I tried to solve for f by using 1/di+1/do=1/f all in terms of f given in the equation

The object is 15 cm from f, so do=f+15 or f-15
The image is 5x that far away, so 5(f+15) or 5(f-15)
And the focus is f

Neither of these worked. I got -/+75cm for f.

I understand the second and third question. For the possible do 15 cm closer to the mirror than f, the image will be virtual and behind the mirror. This is the lesser distance. For the possible do 15 cm further away, the image will be real and in front of the mirror.

2. Aug 31, 2010

### Chi Meson

Draw the situation. Remember that the image distance is measured from the mirror surface, so the distance from the image to the focal point will be di+f, if it is a virtual image, or di -f if real.

3. Sep 1, 2010

### rl.bhat

Try the image disturbance di = f + 5*15
Then

1/f = 1/(f+15) + 1/(f + 75)

Now solve for f.